Number of zeros after convolution

Question

Suppose we have two continous functions $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ that have $n$ and $k$ zeros respectifly. Let $h=f*g$ (convolution) with $m$ number of zeros. Can we say someting about the relationship between $ m$ and $(n,k)$? For exmaple, is it true that $m \le n+k$?

Answer 1

Without further requirements there is no such simple relation.

Case 1: take $f$ integrable with zero integral (you can do this with $f$ taking the value zero just once), and $g=1$. Then $f*g=0$, therefore $m=\infty,n=1,k=0$.

Case 2: take $f$ with compact support, non-zero integral, and $g=1$. Then $f*g=\|f\|_1>0$, therefore $m=0,n=\infty, k=0$.

The only relation that comes to my mind is the following: if both $f$ and $g$ never take the value zero, then neither does $f*g$, because $f$ and $g$ do not change sign and because a positive (resp. negative) function has positive (resp. negative) integral.

EDIT: You can have $m=2d+1,n=1,k=0$, where $d\in\mathbb N$. I'll sketch the construction.

1) Consider the function $f=\chi_{(-1,0)}-\chi_{(0,1)}$ (this has an infinite number of zeros, but we will modify it later). Take $g$ of the form $$g(x)=1+\sum_{i=1}^{d}\chi_{\left(0,1\right)}(x-10i)$$ which is made of $d$ bumps, separated by a distance $10$ (this particular value is not important, it's just so that when taking the convolution with $f$, if one bump "interact" with $f$ the others don't). The $1$ in the front is to make $g$ strictly positive, and doesn't change the convolution with $f$ because the latter has zero integral.

2) The convolution $f*g$ will be $-1$ at the points $-(10i+\frac12)$ and $+1$ at the points $-(10i-\frac12)$, with $i=1,\ldots, d$. By continuity between them there must be a zero.

3) Now modify both $f$ and $g$ a bit, making them continuous and making $f$ positive. For example take the convolution (!) with a kernel $\eta_\varepsilon\in C(\mathbb R)$, built starting from $\eta\in C(\mathbb R)$ with integral one, strictly positive, symmetric, strictly radially decreasing, and setting $\eta_\varepsilon(x)=\frac{1}{\varepsilon}\eta(\frac{x}{\varepsilon})$.

If $\varepsilon$ is sufficiently small all the desired properties will still hold, but now $f$ will be zero only at the origin.